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2013济南中考数学试题原版试题原版答案

发布时间:2013-11-11 09:34:54  

绝密★启用前

济南市2013年初三年级学业水平考试

数 学 试 题

本试题分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.第Ⅰ卷共2页,满分为45分;第Ⅱ卷共6页,满分为75分.本试题共8页,满分为120分.考试时间为120分钟.答卷前,请考生务必将自己的姓名、准考证号、座号、考试科目涂写在答题卡上,并同时将考点、姓名、准考证号、座号填写在试卷规定的位置.考试结束后,将本试卷和答题卡一并交回.本考试不允许使用计算器.

第I卷(选择题 共45分)

注意事项:

第Ⅰ卷为选择题,每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑;如需改动,用橡皮擦干净后,再选涂其他答案标号.答案写在试卷上无效.

一、选择题(本大题共15个小题,每小题3分,共45分.在每小题给出的四个选项中,只有一项是符合题目要求的.)

1.-6的相反数是

11 A.? B. C.-6 D.6 66

2.下图是由3个相同的小立方体组成的几何体,它的主视图是

A. D.

B. C.

3.十八大以来,我国经济继续保持稳定增长,2013年第一季度国内生产总值约为118900亿元,将数字118900用科学记数法表示为

A.0.1189×106 B.1.189×105 C.11.89×104 D.1.189×104

4.如图,直线a,b被直线c所截,a∥b,∠1=130°,则∠2的度数是 c A.130° B.60° C.50° D.40°

5.下列各式计算正确的是 a 2224 A.(a)?a B.a?a?a

222b a2?a8 C.3a?a?2a D.a4?第4题图 ?3x?1>56.不等式组?的解集在数轴上表示正确的是 ?2x≤6

D.

C.

A.

7.为了解七年级学生参与家务劳动的时间,李老师随机调查了七年级8名学生一周内参与家务劳动的时间(单位:小时)分别是:1,2,3,3,3,4,5,6.则这组数据的众数是

A.2.5 B.3 C.3.375 D.5

数学试题 第1页(共16页)

2x6,其结果是 ?x+3x?3

A.2 B.3

C.x+2 D.2x+6

9.如图,在平面直角坐标系中,△ABC的三个顶点的坐标分别 为A(-1,0),B(-2,3),C(-3,1).将△ABC绕点A按 顺时针方向旋转90°,得到△AB?C?,则点B?的坐标为

A.(2,1) B.(2,3)

C.(4,1) D.(0,2)

10.如图,AB是?O的直径,C是?O上一点,AB=10,AC=6,

OD?BC,垂足为D,则BD的长为

A.2 B.3

C.4 D.6

11.已知x2?2x?8?0,则3x2?6x?188.计算

A.54 第10题图 C.-10 12.面,然后将绳子末端拉到距离旗杆离地面2mA.12m C.16m 13.如图,平行四边形OABC的顶点B坐标为(3,0),点D为边AB的图象经过C,D两点,若∠COA= A.8sin2α C.4tanα 14.已知直线l1∥l2∥l3∥l4为h,矩形ABCD方式如图所示,AB=4,BC=62A. 34C. 315.如图,二次函数y?ax2?bx?cx轴交点的横坐标分别为x1,x2,且下列结论正确的是 ..

A.a?0

bC.?>1 2a第15题图

数学试题 第2页(共16页)

绝密★启用前

济南市2013年初三年级学业水平考试

数 学 试 题

第Ⅱ卷(非选择题 共75分)

注意事项:

1.第Ⅱ卷为非选择题,请考生用蓝、黑色钢笔(签字笔)或圆珠笔直接在试卷上作答. 2.答卷前,请考生先将考点、姓名、准考证号、座号填写在试卷规定的位置.

二、填空题(本大题共6个小题,每小题3分,共18分.把答案填在题

中的横线上.)

16.计算:3(2x?1)?6x=. 17.分解因式:a2?4=18.小明和小华做投掷飞镖游戏各5次,两人

成绩(单位:环)如图所示,根据图中的 信息可以确定成绩更稳定的是________

(填“小明”或“小华”).

19.如图,AB是⊙O的直径,点D在⊙O作⊙O 的切线交AB的

延长线于点C,则∠C=.

第18题图

21题图

20.若直线y?kx与四条直线x=1,x=2,y=1,y=2围成的正方形有公共点,则k的取值

范围是 .

21.如图, D,E分别是△ABC边AB,BC上的点,AD=2BD,BE=CE,设△ADF的面

积为S1,△CEF的面积为S2,若S?ABC?6,则S1?S2的值为.

数学试题 第3页(共16页)

三、解答题(本大题共7个小题,共57分.解答应写出文字说明、证明过程或演算步骤.) 22.(本小题满分7分)

01)?tan45°. (1

)计算:

(2)解方程:

23.(本小题满分

7分)

(1)如图,在△ABC和△DCE中,AB∥DC,AB=DC,BC=CE,且点B,C,E在一条直线上. 求证:∠A=∠D.

B E C

第23(1)题图

(2)如图,在矩形ABCD中,对角线AC,BD相交于点O,AB=4,∠AOD=120°, 求AC的长.

A D

C 第23(2)题图

32. ?xx?1

数学试题 第4页(共16页)

24.(本小题满分8分)

某寄宿制学校有大、小两种类型的学生宿舍共50间,大宿舍每间可住8人,小宿舍每间可住6人.该校360名住宿生恰好住满这50间宿舍.求大、小宿舍各有多少间? ....

25.(本小题满分8分)

在一个不透明的袋子中,装有2个红球和1个白球,这些球除了颜色外都相同.

(1)搅匀后从中随机摸出一球,请直接写出摸到红球的概率;

(2)如果第一次随机摸出一个小球(不放回),充分搅匀后,第二次再从剩余的两球中随机摸出一个小球,求两次都摸到红球的概率.(用树状图或列表法求解)

数学试题 第5页(共16页)

26.(本小题满分9分)

如图,点A的坐标是(-

2,0),点B的坐标是(6,0),点C在第一象限内且△OBC为等边三角形,直线BC交y轴于点D,过点A作直线AE⊥BD,垂足为E,交OC于点F.

(1)求直线BD的函数表达式;

(2)求线段OF的长;

(3)连接BF,OE,试判断线段BF和OE的数量关系,并说明理由.

数学试题x 第第6页(共16页)

27.(本小题满分9分)

如图1,在△ABC中,AB=

AC=4,∠ABC=67.5°,△ABD和△ABC关于AB所在的直线对称,点M为边AC上的一个动点(不与点A,C重合),点M关于AB所在直线的对称点为N,△CMN的面积为S.

(1)求∠CAD的度数;

(2)设CM=

x,求S与x的函数表达式,并求x为何值时S的值最大?

(3)S的值最大时,过点C作EC⊥AC交AB的延长线于点E,连接EN(如图2).P为线段EN上一点,Q为平面内一点,当以M,N,P,Q为顶点的四边形是菱形时,请直接..写出所有满足条件的NP的长. ..

第27题图1

N N 第27题图2 A

数学试题 第7页(共16页)

28.(本小题满分9分)

2如图1,抛物线y??x2?bx?c与x轴相交于点A,C,与y轴相交于点B,连接AB,3BC,点A的坐标为(2,0),tan∠BAO=2.以线段BC为直径作⊙M交AB于点D.过点B 作直线l∥AC,与抛物线和⊙M的另一个交点分别是E,F.

(1)求该抛物线的函数表达式;

(2)求点C的坐标和线段EF的长;

(3)如图2,连接CD并延长,交直线l于点N.点P,Q为射线NB上的两个动点(点P在点Q的右侧,且不与N重合),线段PQ与EF的长度相等,连接DP,CQ,四边形CDPQ的周长是否有最小值?若有,请求出此时点P的坐标并直接写出四边形CDPQ周长的最小值;......若没有,请说明理由.

数学试题 第8

绝密★启用前

济南市2013年初三年级学业水平考试

数学试题参考答案及评分意见

二、填空题

1

16.3 17.(a+2)(a-2) 18. 小明 19.20 20.≤k≤2 21.1

2

三、解答题

1)?tan45°22.(1)解:

=1+1 ················································································································ 2分

=2 ···················································································································· 3分 (2)解:去分母,得

3(x?1)=2x ····································································································· 5分

解得 x=3 ······································································································· 6分 检验:把x=3代入原方程,左边=1=右边

∴x=3是原方程的解 ························································································ 7分 23.(1)证明:∵AB∥DC

∴∠B=∠DCE ································································································ 1分 又∵AB=DC,BC=CE

∴△ABC≌△DCE ·························································································· 2分 ∴∠A=∠D ····································································································· 3分

(2)解:∵四边形ABCD是矩形

∴OA=OB=OC=OD ··························································································· 4分 又∵∠AOD=120°

∴∠AOB=60°

∴△AOB为等边三角形 ····················································································· 6分 ∴AO=AB=4

∴AC=2AO=8 ······································································································ 7分

24.解法一:设大宿舍有x间,小宿舍有y间,根据题意得 ················································ 1分

数学试题 第9页(共16页)

?x?y?50 ········································································································· 5分 ?8x?6y?360?

?x?30解方程组得? ··································································································· 7分 y?20?

答:大宿舍有30间,小宿舍有20间.······································································ 8分 解法二:设大宿舍有x间,则小宿舍有(50-x)间,根据题意得 ································· 1分

8x+6(50-x)=360 ·································································································· 5分

解方程得x=30

∴50-x=20 (间) ·································································································· 7分

答:大宿舍有30间,小宿舍有20间. ············································································· 8分

25.解:(1)P(红球)= 2 ········································································································· 2分 3

开 始 (2)解:所有可能出现的结果如图所示:

(一) 红1 红2 白 1 红1 (二) 红2 白

或所有可能出现的结果如下表所示:

红1

红2

白 红1 (红2,红1) (白 ,红1) 红2 红2 (红1,红2 ) (白 ,红2) 白 (红1,白) (红2,白)

··················································································································································· 6分

总共有6种结果,每种结果出现的可能性相同,其中两次都摸到红球的有2种,

∴P(两次都摸到红球)= 21? ········································································ 8分 63

26.解:(1)∵△OBC是等边三角形

∴∠OBC=∠BOC=∠OCB =60°,OB=BC=CO

∵B(6,0)

∴BO=6

∴OD?OB?tan600?

∴点D的坐标为(0, ················································································ 1分

数学试题 第10页(共16页)

设直线BD的表达式为y?kx?b ?6k?b?0∴? ······································································································· 2分 ?b?63

??k??3∴? ??b?63

∴直线BD的函数表达式为y =-x+63 ······················································ 3分

(2)解法一:

∵A (-2,0)

∴AO=2

∵AE⊥BD,∠OBC =60°

∴∠EAO=30° ········································································································· 4分 又∵∠BOC=60°

∴∠AFO=30° ········································································································· 5分 ∴∠OAF=∠OFA

∴OF=AO=2 ········································································································· 6分 解法二:

∵A (-2,0)

∴AO=2

∵OB=OC=BC=6,OA=2

∴AB=8

∵AE⊥BD,∠OBC =60°

∴∠BAE=30°

∴BE=4 ·············································································································· 4分 ∴CE=BC-BE=6-4=2 ∴CF?CE2·········································································· 5分 ??4 ·cos∠ECFcos600

∴OF=OC-CF=6-4=2 ······························································································ 6分

(3)BF=OE ··············································································································· 7分 解法一:

∵A(-2,0),B(6,0)

数学试题 第11页(共16页)

∴AB=8

∵∠CBO=60°,AE?BD

∴∠EAB =30°

∴EB=4

∵CB=6

∴CE=2

∵OF=2

∴CE=OF ·················································································································· 8分 又∵∠OCE=∠BOF=60°,CO=BO

∴△COE ≌△OBF

∴OE=BF ···················································································································· 9分 解法二:

过点E作EG?AB,垂足为G. ∵A(-2,0),B(6,0)

∴AB=8

∵∠CBO=60°,AE?BD

∴∠EAB =30°

∴EB=4

∵CB=6

∴CE=2

在Rt△EGB和Rt△CEF中易求 EG=2,EF=23

EB=4,GB=2,OG=4

在Rt△EGO和Rt△FEB中,由勾股定理得 OE

2 ·························································································· 8分 BF=EF2?

EB2=∴OE=BF ······················································································································ 9分 ( 注:此题解法多样,请阅卷老师根据答题情况合理赋分.)

27.解: (1) ∵AB=AC,∠ABC=67.5°

∴∠ABC=∠ACB=67.5°

∴∠CAB=45° ····································································································· 2分 ∵△ABD和△ABC关于AB所在直线对称

∴∠BAD=∠CAB=45°

∴∠CAD=90° ····································································································· 3分

(2)由(1)可知AN⊥AM

∵点M,N关于AB所在直线对称

∴AM=AN

数学试题 第12页(共16页)

∵CM=x,∴AN =AM=4-x

11∴S=CM?AN=x(4?x) 22

1∴S=?x2?2x ······························································································ 5分 2

∴当x=?2

12?(?)2=2时,S有最大值 ······························································· 6分

(3) NP······································································································ 7分 1?····································································································· 8分 NP2?···································································································· 9分 NP3?28.解:(1)∵点A(2,0),tan∠BAO=2

∴AO=2,BO=4

∴点B的坐标为(0,4) ··················································································· 1分 ?抛物线y=?22x+bx+c过点A,B 3

?8???2b?c?0∴?3 ························································································· 2分 ??c?4

2?b???解得?3

??c?4

∴此抛物线的解析式为y=?

(2)解法一:在图1中连接CF,

令y=0,即?222x?x+4 ······················································ 3分 33222x?x+4=0 33

解得x1??3,x2?2

∴点C坐标为(-3,0),CO=3 ········································································· 4分 令y=4,即?222x?x+4=4 33

解得x1?0,x2??1 ∴点E坐标为(-1,4) ∴BE=1 …………………………………5分 ∵BC为⊙O直径 ∴∠CFB=90°

又∵BO⊥AC ,l∥AC 数学试题 第13页(共16

∴BO⊥l

∴∠FBO=∠BOC=90°

∴四边形BFCO为矩形

∴BF=CO=3

∴EF=BF-BE=3-1=2……………………6分

解法二:

1∵抛物线对称轴为直线x=? 2

∴点A的对称点C的坐标为(-3,0) ···································································· 4分 点B的对称点E的坐标为(-1,4) ···································································· 5分 ∵BC是⊙M的直径

3∴点M的坐标为(?,2) 2

如图1,过点M作MG⊥FB,则GB=GF

∵M(?

∴BG= 3,2) 23 2∴BF=2BG=3

∵点E的坐标为(-1,4),BE=1

∴EF=BF-BE=3-1=2 ············································

(3)四边形CDPQ的周长有最小值. ·······················

理由如下:

∵BC?5,AC?OC∴AC=BC

∵BC为⊙M直径

∴∠BDC=90°,即CD⊥AB

∴D为AB中点

∴点D的坐标为(1,2)

作点D关于直线l的对称点D1(1,6),点C向右平移2个单位得点C1(-1,0), 连接C1 D1与直线l交于点P,点P向左平移两个单位得点Q,四边形CDPQ即为 周长最小的四边形.

解法一:设直线C1D1的函数表达式为y?mx?n

数学试题 第14页(共16页)

??m?n?0?m?3∴?∴? m?n?6n?3??

∴直线C1D1的表达式为 y?3x?3

∵yp?4

∴xp?1 3

1∴点P的坐标为(,4) ······························································································ 8分 3

解法二:直线D1D交直线l于点H,

交 x轴于点K,易得D1K⊥C1K,D1H⊥PH,

由题意可知D1H=2 ,D1K=6,C1K=2,

由直线l∥x轴,易证△D1PH∽△D1 C1K

PHD1H?∴ C1KD1K

∴PH=2 3

21= 33∴BP=BH-PH=1-1∴点P的坐标为(,4) ·································

3

C四边形CDPQ最小=2 ···················

注:本试卷解答题的其他正确解法,请参照上述参考答案及评分意见酌情赋分.

数学试题 第15页(共16页)

数学试题第16页(共16页)

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